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30 January, 06:09

How many molecules of F2 react with 66.6 g NH3? (stoichiometry)

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  1. 30 January, 08:40
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    5.89 * 10^23 molecules of F₂

    Explanation:

    The equation for the reaction between fluorine (F₂) and ammonia (NH₃) is given by;

    5F₂ + 2NH₃ → N₂F₄ + 6 HF

    We are given 66.6 g NH₃

    We are required to determine the number of fluorine molecules

    Step 1: Moles of Ammonia

    Moles = Mass : Molar mass

    Molar mass of ammonia = 17.031 g/mol

    Moles of NH₃ = 66.6 g : 17.031 g/mol

    = 3.911 moles

    Step 2: Moles of Fluorine

    From the equation 5 moles of Fluorine reacts with 2 moles of ammonia

    Therefore,

    Moles of fluorine = Moles of Ammonia * 5/2

    = 3.911 moles * 5/2

    = 9.778 moles

    Step 3: Number of molecules of fluorine

    We know that 1 mole of a compound contains number of molecules equivalent to the Avogadro's number, 6.022 * 10^23 molecules

    Therefore;

    1 mole of F₂ = 6.022 * 10^23 molecules

    Thus,

    9.778 moles of F₂ = 9.778 moles * 6.022 * 10^23 molecules/mole

    = 5.89 * 10^23 molecules

    Therefore, the number of fluorine molecules needed is 5.89 * 10^23 molecules
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